56 lines
2.6 KiB
Markdown
56 lines
2.6 KiB
Markdown
#start of lecture 11
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last lecture we did cauchy euler equations:
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$ax^2{\frac{d^2y}{dx^2}+bx{\frac{ dy }{ dx }}+cy=f(x)},\ x>0$
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where $a,\ b,\ c$ are still constants and $\in \mathbb{R}$
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1) $x=e^t$
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$a{\frac{d^2y}{dt^2}}+(b-a){\frac{dy}{dt}}+cy=f(e^t)$ <- lousy notation, the y here isnt quite the same as in the above definition.
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2) $y=x^r$
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$ar^2+(b-a)r+C=0$
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three cases:
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(i) $r_1\ne r_{2}$
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then: $y_{h}(x)=c_{1}x^{r_{1}}+c_{2}x^{r_{2}}$
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(ii) $r_{1}=r_{2}=r$
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then: $y_{h}(x)=c_{1}x^r+c_{2}x^r\ln(x)$
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(iii) $r_{1,2}=\alpha+i\beta$
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then: $y_{h}(x)=x^2(c_{1}\cos(\ln \beta x)+c_{2}\sin \ln(\beta x))$
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now find one particular solution for a non homogenous soultion, using variation of parameters, combine the y_h and y_p to get y(x).
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not all equations can fall into cauchy euler type.
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$y''+p(x)y'+q(x)y=f(x)$ (1) <- no general solution procudure always
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but, if $y_{1}(x)$ solves $y''+p(x)y'+q(x)y=0$
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then we can find the general solution to the non homogenous equation (1) by guessing it in the form $y(x)=v(x)y_{1}(x)$
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$y'=v'y_{1}+vy_{1}'$
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$y''=v''y_{1}+2v'y_{1}'+vy_{1}''$
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$(v''y_{1}+2v_{1}'y_{1}'+y_{1}''v)+p(x)(v'y_{1}+vy_{1}')+q(x)vy_{1}=f(x)$
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$v\cancelto{ 0 }{ (y_{1}''+p(x)y_{1}'+q(x)y_{1}) }+v''y_{1}+(2y_{1}'+p(x)y_{1})v'=f$
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$y_{1}v''+()$
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$v''+\left( \frac{2y_{1}'}{y_{1}}+p \right)v'=\frac{f}{y_{1}}$
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$v'=u$
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$u'+\left( \frac{2y_{1}'}{y_{1}}+p \right)u=\frac{f}{y_{1}}$<- this is a linear first order equation
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how to solve linear first order equation? we compute the integrating factor $\mu$
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$\mu=e^{\int(2y_{1}'/y_{1}+p)dx}=e^{\ln(y_{1})^2}e^{\int P(x) \, dx}=y_{1}^2e^{\int p(x) \, dx}$
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isnt this nice? some kind of magic. We made some guesses and we arrived somewhere.
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#ex find the general solution to the equation:
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$y''+4xy'+(4x^2+2)y=8e^{-x(x+2)}$
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if $y_{1}(x)=e^{-x^2}$ is one solution.
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therefore were finding the solution of the form: $y(x)=v(x)y_{1}=v(x)e^{-x^2}$
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$v'=u$
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$u'+\left( \frac{2y_{1}'}{y_{1}}+4x \right)u=\frac{8{e^{-x^2}e^{-2x}}}{e^{-x^2}}$ <-(p(x)=4x)
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$u'+\left( \frac{2{e^{-x^2}(-2x)}}{e^{-x^2}}+4x \right)u=8e^{-2x}$
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$u'=8e^{-2x}$
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$u=-4e^{-2x}+c_{1}$
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$v'=u=-4e^{-2x}+c_{1}$
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$v(x)=2e^{-2x}+c_{1}x+c_{2}$
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general solution:
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$$y(x)=v(x)y_{1}(x)=(2e^{-2x}+c_{1}x+c_{2})e^{-x^2}$$
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## Free vibrations
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$mr^2+br+k=0$ characteristic polynomail
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(i) $r_{1}\ne r_{2}$ $b^2-4mk>0$
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$y_{h}(t)=c_{1}e^{r_{1}t}+c_{2}e^{r_{2}t}$
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$r_{1,2}=-\frac{b}{2m}\pm \frac{\sqrt{ b^2-4mk }}{2m}<0$
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then the limit of the homogenous solution is 0 as t->$\infty$ (over damped case)
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(ii) $r_{1}=r_{2}=-\frac{b}{2m}$
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$r_{1}=r_{2}=-\frac{b}{2m}$
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$y_{h}(t)=e^-\frac{b}{2m}+c_{2}te^{-b/2m}t$ limit =0 as t approches inf critically damped |